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405. Convert a Number to Hexadecimal

Description

Given an integer num, return a string representing its hexadecimal representation. For negative integers, two’s complement method is used.

All the letters in the answer string should be lowercase characters, and there should not be any leading zeros in the answer except for the zero itself.

Note: You are not allowed to use any built-in library method to directly solve this problem.

 

Example 1:

Input: num = 26
Output: "1a"

Example 2:

Input: num = -1
Output: "ffffffff"

 

Constraints:

• -231 <= num <= 231 - 1

Solutions

Solution 1: Bit Manipulation

This implementation uses bit manipulation. It traverses the relevant values and updates its state as each value is processed. After all required states have been considered, the maintained result is returned.

Solution 2

This implementation uses bit manipulation. It traverses the relevant values and updates its state as each value is processed. After all required states have been considered, the maintained result is returned.

• Java
• C++
• Python
• Go

• class Solution {
      public String toHex(int num) {
          if (num == 0) {
              return "0";
          }
          StringBuilder sb = new StringBuilder();
          while (num != 0) {
              int x = num & 15;
              if (x < 10) {
                  sb.append(x);
              } else {
                  sb.append((char) (x - 10 + 'a'));
              }
              num >>>= 4;
          }
          return sb.reverse().toString();
      }
  }
  
  
  // Solution 2
  class Solution {
      public String toHex(int num) {
          if (num == 0) {
              return "0";
          }
          StringBuilder sb = new StringBuilder();
          for (int i = 7; i >= 0; --i) {
              int x = (num >> (4 * i)) & 0xf;
              if (sb.length() > 0 || x != 0) {
                  char c = x < 10 ? (char) (x + '0') : (char) (x - 10 + 'a');
                  sb.append(c);
              }
          }
          return sb.toString();
      }
  }
  
  

• class Solution {
  public:
      string toHex(int num) {
          if (num == 0) return "0";
          string s = "";
          for (int i = 7; i >= 0; --i) {
              int x = (num >> (4 * i)) & 0xf;
              if (s.size() > 0 || x != 0) {
                  char c = x < 10 ? (char) (x + '0') : (char) (x - 10 + 'a');
                  s += c;
              }
          }
          return s;
      }
  };
  

• class Solution:
      def toHex(self, num: int) -> str:
          if num == 0:
              return '0'
          chars = '0123456789abcdef'
          s = []
          for i in range(7, -1, -1):
              x = (num >> (4 * i)) & 0xF
              if s or x != 0:
                  s.append(chars[x])
          return ''.join(s)
  
  

• func toHex(num int) string {
  	if num == 0 {
  		return "0"
  	}
  	sb := &strings.Builder{}
  	for i := 7; i >= 0; i-- {
  		x := num >> (4 * i) & 0xf
  		if x > 0 || sb.Len() > 0 {
  			var c byte
  			if x < 10 {
  				c = '0' + byte(x)
  			} else {
  				c = 'a' + byte(x-10)
  			}
  			sb.WriteByte(c)
  		}
  	}
  	return sb.String()
  }
  

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